# Parallel Lines in a Cyclic Quadrilateral

## What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The applet attempts to suggest a problem from the 1993-1994 Saint Petersburg (Russia) Regional Mathematical Olympiad for grade 9

Two crossing chords AC and BD are drawn in a given circle. M is a point on AC such that

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Let G be the intersection of AC and BD. We shall focus on a pair of similar triangle: AGD and BGC. Importantly, two inscribed angles are equal -

Triangles AGD and BGC are similar because

AG/AD = BG/BC.

Since AD = AM and BC = BN, the proportion could be rewritten as

AG/AM = BG/BN.

Triangles ABG and MNG have two pairs of proportional sides and equal angles in-between. They are therefore similar and have pairwise equal angles. The lines AB and MN are parallel because, say, equal angles GMN and GAB are, depending on the configuration, either corresponding or alternate with respect to the transversal AM.

Note that the requirement that the chords AD and BC cross is redundant. Both the problem and the solution stand even if they do not.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny